From the top of a tower of height 50 m a ball is projected upwards with a speed of 30 m/s at an angle of 30° to the horizontal. Take g = 10 m/s². Find the distance from the foot of the tower where the ball hits the ground (m).
Physics · Kinematics part-2 Projectile Motion · Projectile from height
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Question

From the top of a tower of height 50 m a ball is projected upwards with a speed of 30 m/s at an angle of 30° to the horizontal. Take g = 10 m/s². Find the distance from the foot of the tower where the ball hits the ground (m).

Physics · Kinematics part-2 Projectile Motion · Projectile from height

G1 Physics · Accepted Answer

Vertical component of projection is uᵧ = 30 sin30° = 15 m/s. Taking the launch point as y = 0 and the ground as y = -50 m: -50 = 15t - 5t², giving t = 5 s. The horizontal component is uₓ = 30 cos30° = 15√3 m/s. Hence horizontal distance from the foot of the tower = uₓt = 75√3 ≈ 129.90 m.